Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in PYTHON):

Solution:
from __future__ import division
import sys
range = xrange
input = sys.stdin.readline

sy,a,b = [int(x) for x in input().split()]

n = int(input())

inp = [int(x) for line in sys.stdin for x in line.split()]
ii = 0

L = []
R = []
for _ in range(n):
    l = inp[ii]
    r = inp[ii+1]
    ii+=2
    L.append(l)
    R.append(r)

Lsum = [0]
for l in L:
    Lsum.append(Lsum[-1]+l)
Rsum = [0]
for r in R:
    Rsum.append(Rsum[-1]+r)

out = []
q = inp[ii]
ii+=1
for _ in range(q):
    px = inp[ii]
    py = inp[ii+1]
    ii+=2

    shade = 0
    

    # L' = px + (l-px)/py*(py-sy)
    # R' = px + (l-px)/py*(py-sy)
    
    A = 0
    B = len(L)
    while A<B:
        M = (A+B)//2
        l = L[M]
        l = px + (l-px)/py*(py-sy)
        if l<a:
            A = M+1
        else:
            B = M
    ind1 = A

    if ind1==n:
        lista = [ind1-1]
    else:
        A = ind1-1
        B = len(R)-1
        while A<B:
            M = (A+B+1)//2
            r = R[M]
            r = px + (r-px)/py*(py-sy)
            if b<r:
                B = M-1
            else:
                A = M
        ind2 = A

        if ind2==ind1-1:
            lista = [ind1-2,ind1-1,ind1,ind1+1]
        else:
            
            l = Lsum[ind2+1]-Lsum[ind1]
            r = Rsum[ind2+1]-Rsum[ind1]
            
            l = (l)/py*(py-sy)
            r = (r)/py*(py-sy)

            shade += r-l

            lista = [ind1-1,ind2+1]

    
    for i in lista:
        if i<0 or i>=n:
            continue
        l = L[i]
        r = R[i]
        l = px + (l-px)/py*(py-sy)
        r = px + (r-px)/py*(py-sy)

        if l<=a:
            l = a
        if b<=r:
            r = b
        shade += max(r-l,0)
    out.append('%.10f'%(shade))
print '\n'.join(out)